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Math-5536-A3 · revision r1 · on Math-5536

gpt pro 5.4 attempt on Math-5536

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Contents
  1. Erdős Problem #261
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-2)
  5. 4) ATTACK PLAN (ROUND-2)
  6. 5) WORK (ROUND-2)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL (EXACTLY ONE)
  9. 8) COMPLETION ESTIMATE
  10. 9) REFERENCES

Erdős Problem #261

1) ROUND-2 OBJECTIVE

I pursue path (C) in the fail-safe sense: isolate the exact finite-combinatorial structure left open by Round 1, and exploit it to produce new infinite families beyond the Cusick--Borwein--Loring family already proved in Round 1.

2) Round-1 FOUNDATION USED

From Round 1 I use only the already-proved contiguous-block identity: \[ \frac{n}{2^n}=\sum_{k=n+1}^{n+m}\frac{k}{2^k} \qquad\text{for}\qquad n=2^{m+1}-m-2. \] This is the Borwein--Loring realization of Erdős' remark about Cusick's argument.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is an exact offset parametrization of every solution.

If \[ \frac{n}{2^n}=\sum_{i=1}^{t}\frac{a_i}{2^{a_i}} \qquad (t\geq 2,\ a_1<\cdots<a_t), \] then necessarily every $a_i>n$, so after writing \[ a_i=n+d_i \qquad (1\leq d_1<\cdots<d_t), \] the equation becomes \[ n=\sum_{i=1}^{t}\frac{n+d_i}{2^{d_i}}. \] Equivalently, if $D=\{d_1,\dots,d_t\}$ then \[ n\left(1-\sum_{d\in D}2^{-d}\right)=\sum_{d\in D} d\,2^{-d}. \] Thus \[ n=\frac{\sum_{d\in D} d\,2^{-d}}{1-\sum_{d\in D}2^{-d}}. \tag{261.1} \] This is a necessary and sufficient condition for solvability.

4) ATTACK PLAN (ROUND-2)

The exact gap left by Round 1 was this: we had one explicit infinite family, but no structural reduction of the full problem. The plan is now:

  1. prove the offset parametrization $261.1$;
  2. use it to reinterpret the problem as a finite-subset existence problem;
  3. build new infinite families by choosing $D$ of the special form \[ D=\{1,2,\dots,M\}\setminus\{M-r,M\}. \]

This overcomes the earlier gap because it turns a problem about arbitrary exponents $a_i$ into a problem about finite binary-offset sets.

5) WORK (ROUND-2)

Step 1: every exponent lies strictly above $n$.

Let \[ f(m):=\frac{m}{2^m}\qquad (m\geq 1). \] On positive integers, $f$ is nonincreasing, with the only tie $f(1)=f(2)=1/2$, and it is strictly decreasing for $m\geq 2$.

Suppose \[ \frac{n}{2^n}=\sum_{i=1}^{t}\frac{a_i}{2^{a_i}} \] with $t\geq 2$ and distinct $a_i$. If some $a_i\leq n$, then \[ \frac{a_i}{2^{a_i}}\geq \frac{n}{2^n}, \] and the inequality is strict unless we are in one of the obvious one-term equality cases. Since all summands are positive and $t\geq 2$, this is impossible. Therefore \[ a_i>n\qquad\text{for all }i. \tag{261.2} \]

Step 2: exact offset parametrization.

Define $d_i:=a_i-n$, so by $261.2$ each $d_i$ is a positive integer. Then \[ \frac{n}{2^n} = \sum_{i=1}^{t}\frac{n+d_i}{2^{n+d_i}}. \] Multiplying by $2^n$ gives \[ n=\sum_{i=1}^{t}\frac{n+d_i}{2^{d_i}} = n\sum_{i=1}^{t}2^{-d_i}+\sum_{i=1}^{t} d_i\,2^{-d_i}. \] Hence \[ n\left(1-\sum_{i=1}^{t}2^{-d_i}\right)=\sum_{i=1}^{t} d_i\,2^{-d_i}, \] and therefore \[ n=\frac{\sum_{i=1}^{t} d_i\,2^{-d_i}}{1-\sum_{i=1}^{t}2^{-d_i}}. \] Conversely, if $D=\{d_1<\cdots<d_t\}$ is any finite set of positive integers such that the right-hand side of $261.1$ is a positive integer $n$, then setting \[ a_i:=n+d_i \] gives \[ n\left(1-\sum_{i=1}^{t}2^{-d_i}\right)=\sum_{i=1}^{t} d_i\,2^{-d_i}, \] which rearranges back to \[ \frac{n}{2^n}=\sum_{i=1}^{t}\frac{a_i}{2^{a_i}}. \] So $261.1$ is an exact necessary-and-sufficient parametrization.

Step 3: reinterpretation of the open all-$n$ problem.

The question "is there a solution for every $n$?" is now exactly equivalent to:

Equivalent formulation. For every positive integer $n$, does there exist a finite set \[ D\subset \{1,2,3,\dots\} \] such that \[ \sum_{d\in D}2^{-d}<1 \qquad\text{and}\qquad n=\frac{\sum_{d\in D} d\,2^{-d}}{1-\sum_{d\in D}2^{-d}}\, ? \]

Step 4: a new two-omission family.

Fix integers $M>r\geq 1$ and take \[ D=\{1,2,\dots,M\}\setminus\{M-r,M\}. \] Then \[ \sum_{d=1}^{M}\frac{n+d}{2^{n+d}} = \frac{n+2}{2^n}-\frac{n+M+2}{2^{n+M}} \] by the same dyadic-tail identity used in Round 1.

Subtracting the omitted terms gives \[ \sum_{d\in D}\frac{n+d}{2^{n+d}} = \frac{n+2}{2^n} -\frac{n+M-r}{2^{n+M-r}} -\frac{n+M+2}{2^{n+M}} -\frac{n+M}{2^{n+M}}. \] Requiring this to equal $n/2^n$ and multiplying by $2^{n+M}$ yields \[ 2^{M+1} = (2^r+2)n+(2^r+2)M-r\,2^r+2. \] Hence \[ n= \frac{2^{M+1}-(2^r+2)M+r\,2^r-2}{2^r+2}. \tag{261.3} \] Whenever the right-hand side is a positive integer, we get a valid solution \[ \frac{n}{2^n} = \sum_{\substack{1\leq d\leq M\\ d\neq M-r,\ d\neq M}} \frac{n+d}{2^{n+d}}. \]

Step 5: two explicit infinite families.

Family A (new in this round). Take $r=3$. Then $261.3$ becomes \[ n=\frac{2^{M+1}-10M+22}{10} =\frac{2^M-5M+11}{5}. \] This is an integer exactly when \[ 2^M\equiv -11\equiv 4 \pmod{5}, \] i.e. when \[ M\equiv 2\pmod{4}. \] Therefore, for every $M\equiv 2\pmod{4}$, \[ n=\frac{2^M-5M+11}{5} \] has a solution with \[ D=\{1,2,\dots,M\}\setminus\{M-3,M\}. \] The first values are \[ n=9,\ 197,\ 3265,\dots \] For instance, when $M=6$, \[ \frac{9}{2^9} = \frac{10}{2^{10}}+\frac{11}{2^{11}}+\frac{13}{2^{13}}+\frac{14}{2^{14}}. \]

Family B (new in this round). Take $r=4$. Then $261.3$ becomes \[ n=\frac{2^{M+1}-18M+62}{18} =\frac{2^M-9M+31}{9}. \] This is an integer exactly when \[ 2^M\equiv -31\equiv 5 \pmod{9}. \] Since the powers of $2$ modulo $9$ have period $6$, this happens exactly when \[ M\equiv 5\pmod{6}. \] Therefore, for every $M\equiv 5\pmod{6}$, \[ n=\frac{2^M-9M+31}{9} \] has a solution with \[ D=\{1,2,\dots,M\}\setminus\{M-4,M\}. \] The first values are \[ n=2,\ 220,\ 14550,\dots \] For instance, when $M=5$, \[ \frac{2}{2^2} = \frac{4}{2^4}+\frac{5}{2^5}+\frac{6}{2^6}. \]

Remark. The Round 1 Borwein--Loring family is the case $r=1$ in $261.3$, because then \[ D=\{1,2,\dots,M-2\} \] and \[ n=2^{M-1}-M=2^{(M-2)+1}-(M-2)-2. \]

6) ADVERSARIAL VERIFICATION

  • Monotonicity issue at $1,2$. The only tie for $m/2^m$ is $1/2=2/4$. This does not invalidate Step 1, because the right-hand side has at least two positive summands.
  • Necessity versus sufficiency. Formula $261.1$ was proved both ways. No hidden assumption remains.
  • Distinctness of exponents. Distinct $d_i$ give distinct $a_i=n+d_i$ automatically.
  • Size of $t$. In Family A we have $t=M-2\geq 4$ for $M\geq 6$; in Family B we have $t=M-2\geq 3$ for $M\geq 5$.
  • Arithmetic checks. For $r=3$ the congruence condition is exactly $M\equiv 2\pmod 4$; for $r=4$ it is exactly $M\equiv 5\pmod 6$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 gives a full parametrization of all finite solutions and two new explicit infinite families beyond the Round 1 contiguous-block family. The open "for every $n$?" question is now reduced exactly to the finite-subset existence problem encoded by $261.1$.

8) COMPLETION ESTIMATE

COMPLETION: 68%

9) REFERENCES

  • [EP261] T. F. Bloom, Erdős Problem #261, problem page, accessed 2026-03-07.
  • [Er88c] P. Erdős, On the irrationality of certain series: problems and results, 1988; see p. 104.
  • [BoLo90] P. Borwein and L. Loring, Some questions of Erdős and Graham on numbers of the form $\sum g_n/2^{g_n}$, Amer. Math. Monthly 97 $1990$, 22--26.
  • [TUZ20] S. Tengely, M. Ulas, and J. Zygadło, On a Diophantine equation of Erdős and Graham, Revista Colombiana de Matemáticas 54 $2020$, 445--459.